The first part deals with a semilinear elliptic boundary-value problem of the form −∆u=f(u, x) ∈Ω, u= 0 onB0:=∪k−1j=0Γj, u=ρ onB:=∪mj=kΓj, (1.1) where f(u, x

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ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

SEMILINEAR ELLIPTIC BOUNDARY-VALUE PROBLEMS ON BOUNDED MULTICONNECTED DOMAINS

WENJIE GAO, JUNYU WANG, ZHONGXIN ZHANG

Abstract. A semilinear elliptic boundary-value problem on bounded multiconnected domains is studied. The authors prove that under suitable conditions, the problem may have no solutions in certain cases and many have one or two nonnegative solutions in some other cases. The radial solutions were also studied in annular domains.

1. Introduction

This paper consists of two parts. The first part deals with a semilinear elliptic boundary-value problem of the form

−∆u=f(u, x) ∈Ω, u= 0 onB0:=∪^k−1_j=0Γj,

u=ρ onB:=∪^m_j=kΓj,

(1.1)

where f(u, x) ∈ C(R+×Ω;R), R+ := [0,+∞), Ω is a bounded multiconnected domain in Rⁿ, n ≥ 2, Γ0 is its outer boundary, ∪^m_j=1Γj its inner boundary, Γ0, Γ1, . . . ,Γmare all sufficiently smooth closed surfaces so that the Green’s function G(x, y), for−∆ with zero Dirichlet boundary conditions is existent (See, e.g., [2, p.112]), k∈ {1,2, . . . , m}, and the constantρ≥0 is given. Some particular cases of Problem (1.1) were considered by Bandle and Peletier [3], by Lee and Lin [5] and by Hai [4], in whichf ∈(R+;R+) and Ω is a domain with a “hole”.

The second part is devoted to the semilinear elliptic boundary-value problem, namely,

−∆u=f(u,|x|) 0< α <|x|< β <+∞,

u= 0 on|x|=α, u=ρ on|x|=β (1.2) where f ∈C(R+×J;R+), J = [α, β], |x| =p

x²₁+· · ·+x²_n. Problem (1.2) was studied by Lee and Lin [5] and by Hai [4] in the casef ∈C(R+;R+).

We study Problems (1.1) and (1.2) motivated by the following results recently established by Hai [4].

2000Mathematics Subject Classification. 34B15.

Key words and phrases. Semilnear elliptic boundary-value problem; nonnegative solution;

radial solution; nonexistence; existence; multiplicity.

c

2005 Texas State University - San Marcos.

Submitted May 6, 2004. Published January 5, 2005.

Supported by NSFC and by Excellent Young Teacher’s Foundation of MOEC..

1

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Theorem 1.1. Let Ω ⊂ Rⁿ, n ≥ 2, be a bounded domain with a hole and f in C(R+;R+)satisfy

u→0+lim f(u)

u = 0 and lim

u→+∞

f(u)

u = +∞.

Then there exists a positive number ρ^∗ such that Problem (1.1)has a positive solution for ρ∈(0, ρ^∗)and no solution for ρ > ρ^∗.

Theorem 1.2. Let f ∈C(R+;R+)be convex and satisfy lim inf

u→0+

Ru 0 f(s)ds

u² = 0 and lim

u→+∞

f(u)

u = +∞.

Then there exists a positive number ρ^∗ such that Problem (1.2) has at least two positive radial solutions for ρ ∈(0, ρ^∗), at least one for ρ=ρ^∗ and none for any ρ > ρ^∗.

The above results extend and complement the corresponding results in [3, 5].

As in [4], our purpose is to extend, improve and complement the corresponding results in [3, 4, 5]. In the first part, we will prove three theorems. The second theorem extends and complements Theorem 1.1, in which the functionf is allowed to contain variablexand is not necessarily nonnegative, also,f(u, x) is not necessarily superlinear atu= 0 and atu= +∞. In the second part, we obtain a similar result as Theorem 1.2 where the function f may depends on the variable|x| and the convexity constrain tof is removed, alsof is not required to be superlinear at u= 0 andu= +∞.

2. Results in multiconnected domains For the first part, we make the following assumptions:

(A1) f ∈C(R+×Ω;R).

(A2) M := sup{λ1u−f(u, x) :u∈R+, x∈Ω}<+∞.

Here and throughout this section,λ1 denotes the first eigenvalue of the problem

−∆φ=λφ in Ω

φ= 0 on∂Ω =B0∪B. (2.1)

The positive eigenfunction corresponding toλ1 is denoted byφ1(x) withkφ1k= 1, wherek · kstands for the supremum norm.

(A3)

lim sup

u→0+

max{|f(u, x)|:x∈Ω}

u < 1

kgk, where

g(x) :=

Z

Ω

G(x, y)dy, x∈Ω. (2.2)

(A4) f(0, x)≥0 for allx∈Ω and lim sup

u→+∞

u < 1

kgk. Concerning problem (1.1), we can establish the following results.

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Theorem 2.1. Let (A1) and(A2)hold. Then problem (1.1)has no solution for ρ >ρ˜:= MR

Ωφ₁(x)dx λ1

R

Ωφ1(x)h(x)dx. (2.3)

Here h(x)is the harmonic function defined onΩsatisfying

h(x) = 0 on B₀ and h(x) = 1 on B. (2.3) Theorem 2.2. Let (A1),(A2) and(A3)hold. Then there exists a positive number ρ^∗such that problem(1.1)has a nonnegative solution forρ∈[0, ρ^∗)and no solution forρ > ρ^∗.

Theorem 2.3. Let (A1) and (A4) hold. Then problem (1.1) has a nonnegative solution for all ρ ≥ 0. In addition, assume that f(0, x) ≡ 0 for all x ∈ Ω and there exists a δ >0 such that f(u, x)≥λ₁ufor all u∈[0, δ]and all x∈Ω. Then Problem (1.1)withρ= 0 has at least one positive solution.

Theorems 2.1 and 2.3 are new and Theorem 2.2 extends and complements the corresponding theorems in [3, 4, 5]. Before proving these theorems, we make several remarks.

Remark 2.4. A function u∈C(Ω;R) is a solution to Problem (1.1), if and only if it is a solution to the integral equation

u(x) = Z

Ω

G(x, y)f(u(y), y)dy+ρh(x), x∈Ω. (2.4) Clearly, each solution to Problem (1.1) is positive when f ∈C(R+×Ω;R+) and ρ >0.

Remark 2.5. ¿From the maximum principle, we know that 0< h(x)<1 in Ω.

Remark 2.6. According to (2.1) and (2.4), we have 0< φ1(x) =λ1

Z

Ω

G(x, y)φ1(x)dx≤1 in Ω and henceλ₁>1/kgk.

Remark 2.7. It is obvious that Theorems 2.1, 2.2 and 2.3 are still valid if the boundary conditions in (1.1) are replaced byu=ρonB₀,u= 0 onB.

Proof of Theorem 2.1. Suppose to the contrary that Problem (1.1) has a solution, u(x) =

Z

Ω

G(x, y)f(u(y), y)dy+ρh(x) =:w(x) +ρh(x), x∈Ω, for someρ >ρ. In this case, we have˜ −∆w(x) =f(u(x), x) in Ω. Consequently,

ρλ1

Z

Ω

φ1(x)h(x)dx= Z

Ω

λ1φ1(x)u(x)dx− Z

Ω

λ1φ1(x)w(x)dx

= Z

Ω

λ1φ1(x)u(x)dx+ Z

Ω

φ1(x)∆w(x)dx

= Z

Ω

φ1(x)(λ1u(x)−f(u(x), x)dx

≤M Z

Ω

φ1(x)dx,

i.e.,ρ≤ρ, a contradiction. Theorem 2.1 is thus proved.˜

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Proof of Theorem 2.2. To prove Theorem 2.2, we first define a mappingK:E7→E by setting

(Kw)(x) :=

Z

Ω

G(x, y)f^∗(w(y), y)dy+ρh(x), ∀w∈E, (2.5) whereE:=C(Ω;R) and

f^∗(u, x) :=

(f(0, x) ifu <0, f(u, x) ifu≥0.

It is easy to check that K is completely continuous on E. ¿From (A3), we know that there exists an >0 such that

lim sup

u→0+

u < 1

kgk+ < 1 kgk and hence there exists aσ >0 such that

|f(u, x)| ≤ u

kgk+ for allu∈[0, σ] and allx∈Ω, from which it follows that

f(0, x)≡0 for allx∈Ω. (2.6)

Clearly,u(x)≡0 is a trivial solution to Problem (1.1) withρ= 0. Put Dσ:={w∈E:kwk ≤σ} and σ^∗:=σ 1− kgk

kgk+ . Then for each fixedw∈Dσ and each fixedρ∈[0, σ^∗], we have

kKwk ≤ kgk

kgk+σ+σ^∗=σ

which means that K is a compactly continuous mapping fromDσ into itself. The Schauder fixed point theorem tells us thatK has a fixed pointu∈Dσ, i.e.,

−∆u=f^∗(u(x), x) in Ω, u= 0 onB₀, u=ρonB.

¿From (2.6) and the maximum principle, we know thatu(x)≥0 on Ω, which shows that the fixed pointu(x)∈Dσ is also a nonnegative solution to Problem (1.1) with ρ∈[0, σ^∗].

Let

ρ^∗= sup{ρ≥0 : Problem (1.1) has a nonnegative solution}.

¿From the previous results,ρ^∗ ∈[σ^∗,ρ]. We are now going to prove that Problem˜ (1.1) has a nonnegative solution for allρ∈[0, ρ^∗).

For each givenρ∈[0, ρ^∗), from the definition ofρ^∗, we can choose a ¯ρ∈(ρ, ρ^∗) such that Problem (1.1)_ρ_¯ has a nonnegative solution ¯u(x). Let ξ(x) ≡ 0 and η(x) = ¯u(x). Then

−∆ξ(x)≡0≡f(ξ(x), x) ∈Ω, ξ(x) = 0 on B₀, ξ(x) = 0≤ρ onB,

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and

−∆η(x)≡f(η(x), x) in Ω, η(x) = 0 onB0, η(x) = ¯ρ > ρ onB.

i.e., ξ(x) is a lower solution to Problem (1.1) andη(x) is an upper solution. Em- ploying the method of upper and lower solutions, we can find a solution u(x) to Problem (1.1) with

0≡ξ(x)≤u(x)≤η(x) = ¯u(x) onΩ.

The proof of Theorem 2.2 is complete.

Proof of Theorem 2.3. From (A4), we know that there exists an >0 such that lim sup

u→+∞

u ≤ 1

kgk+ < 1 kgk and hence there exists anN >0 such that

|f(u, x)| ≤ u

kgk+ for all u≥N and allx∈Ω.

For each givenρ≥0, we put

Dβ:={w∈E:kwk ≤β}, β:= 1− kgk

kgk+ −1

N+ρ+kgkmax{|f(u, x)|:x∈Ω, 0≤u≤N} . We define a mappingK:E7→E by (2.5). For each fixedw∈Dβ, we have

kKwk<kgk

max{|f(u, x)|: 0≤u≤N, x∈Ω}+ β kgk+

+ρ+N =β, which implies thatKis a compactly continuous mapping from D_β into itself. The Schauder fixed point theorem tells us thatK has a fixed pointu∈D_β, i.e.,

−∆u(x) =f^∗(u(x), x) in Ω, u(x) = 0 onB0, u(x) =ρonB.

From the maximum principle and the assumption that f(0, x) ≥0 for allx ∈Ω, we know that

u(x)≥0 on Ω.

This shows that the fixed pointu∈D_β is a nonnegative solution of (1.1).

We now assume thatf(0, x)≡0 for allx∈Ω and there exists aδ >0 such that f(u, x)≥λ1ufor allu∈[0, δ] and allx∈Ω.

In this case,u(x)≡0 is a trivial solution to Problem (1.1)0. Then we consider the modified boundary-value problem

−∆u(x) = ¯f(u(x), x) in Ω,

u(x) = 0 on∂Ω, (2.7)

where the function

f¯(u, x) :=

(f(δφ1(x), x) ifu < δφ1(x), f(u, x) ifu > δφ1(x)

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satisfies (A1) and (A4) again. From the above discussion, we know that Problem (2.7) has a solutionu∈Dβ. The maximum principle tells us that

u(x)≥δφ₁(x) on Ω.

This shows that the solutionu(x) is also a positive solution to Problem (1.1) with

ρ= 0. Theorem 2.3 is thus proved.

3. Results in annular domains

In this section, we restrict our attention to the multiplicity of radial solutions to (1.2). When we seek for a radial solution to (1.2), the problem can be rewritten as

−(k(t)u⁰(t))⁰=k(t)f(u(t), t), α < t < β,

u(α) = 0, u(β) =ρ, (3.1)

where k(t) = tⁿ⁻¹, n ≥ 2. Concerning Problem (3.1), we make the following hypotheses:

(H1) f ∈C(R+×J;R+),J := [α, β].

(H2)

lim sup

u→0

max{f(u, t) :t∈J}

u < 1

kgk, (3.2)

where

g(t) = Z β

α

G(t, s)ds, t∈J;

G(t, s) = (_k(s)

P(β)(P(β)−P(t))P(s), α≤s≤t≤β,

k(s)

P(β)(P(β)−P(t))P(t), α≤t≤s≤β;

P(t) = Z t

α

dr

k(r), t∈J.

(3.3)

(H3) There exists an interval [a, b] withα < a < b < β, such that lim inf

u→+∞

min{f(u, t) :a≤t≤b}

u > 1

m, where

m=δmax{

Z b a

G(t, s)ds:a≤t≤b}, δ= min{q(t) :a≤t≤b}, q(t) = minP(t)

P(β), P(β)−P(t)

P(β) , t∈J .

(3.4)

(H4) f(u, t) is nondecreasing inu≥0 for each fixed t∈J.

(H4)* f(u, t) is locally Lipschitz continuous inu≥0 for each fixedt∈J.

Theorem 3.1. Let (H1),(H2),(H3) and(H4) (or(H4)*) hold. Then there exists a positive number ρ^∗ such that Problem (1.2) has at least two nonnegative radial solutions for ρ∈[0, ρ^∗), at least one forρ=ρ^∗ and none for anyρ > ρ^∗.

Remark 3.2. Theorem 3.1 is still valid when the boundary conditions in (3.1) are replaced by

u(α) =ρ, u(β) = 0.

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Clearly, Theorem 3.1 is an extension and improvement of the results in [4, The- orem 1.2]. We need the following lemmas.

Lemma 3.3. Letf ∈C(R+×J; R+)andu(t)a nonnegative solution to Problem (3.1). Then

u(t)≥ kukq(t), t∈J .

Proof. Note that Problem (3.1) is equivalent to the integral equation u(t) =

Z β α

G(t, s)f(u(s), s)ds+ρh(t), t∈J, (3.5) where

h(t) :=P(t)/P(β), t∈J. (3.6)

Let kuk = u(t0). Then t0 ∈ (α, β]. If t0 = β, then kuk = ρ and hence u(t) = ρh(t)≥ kukq(t),t∈J. Ift0∈(α, β), then

u(t) = Z β

α

G(t, s)

G(t₀, s)G(t0, s)f(u(s), s)ds+ρh(t0)h(t) h(t₀)

≥q(t) Z β

α

G(t0, s)f(u(s), s)ds+ρh(t0)q(t)

=q(t)kuk for allt∈J.

Here we have used the fact that G(t, s)

G(t₀, s) ≥q(t) for allt, sin (α, β),

the proof of which can be found in [1].

Lemma 3.4. Let(H1)and(H3)hold. Then there exists a positive numberM such that all solutions u(t)to (3.1)satisfy kuk< M.

Proof. By Lemma 3.3, we know that ifu(t) is a nonnegative solution to Problem (3.1), then

u(t)≥δkuk for allt∈[a, b],

where the constantδis defined by (3.4). From (H3), we know that there exists an >0 such that

lim inf

u→+∞

min{f(u, t); t∈[a, b]}

u > 1

m− > 1 m, and hence there exists anM >0 such that

f(u, t)≥ u

m− for allu≥δM and allt∈[a, b]. (3.7) We now claim that kuk < M for all nonnegative solution u(t) to (3.1), where the constant M satisfies (3.7). If the claim is false, then Problem (3.1) has a nonnegative solutionu(t) withkuk ≥M. In this case, we have

u(t)≥ Z b

a

G(t, s)f(u(s), s)ds≥ δkuk m−

Z b a

G(t, s)ds for allt∈[a, b]; i.e.,

kuk ≥ m m−kuk.

This is a contradiction which proves the claim.

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Proof of Theorem 3.1. We first define a mappingK:E7→E by setting (Kw)(t) =

Z β α

G(t, s)f^∗(w(s), s)ds+ρh(t), whereE=C(J;R) and

f^∗(u, t) =

(f(0, t) ifu <0, f(u, t) ifu≥0.

It is easy to check that K is completely continuous on E. ¿From (H2), we know that there exists an >0 such that

lim sup

u→0+

max{f(u, t) :t∈J}

u < 1

kgk+ < 1 kgk and hence there exists aσ >0 such that

0≤f(u, t)≤ u kgk+

for allu∈[0, σ] and allt∈J, which implies that f(0, t)≡0 for allt∈J. We now put

Dσ ={w∈E:kwk ≤σ} and σ^∗=σ 1− kgk kgk+

. Then for each fixedw∈D_σ and each fixedρ∈[0, σ^∗], we have

kKwk ≤ kgk

kgk++σ^∗=σ,

which implies thatKis a completely continuous mapping fromD_σinto itself. The Schauder fixed point theorem tells us thatK has a fixed pointu∈D_σ, i.e.,

−(k(t)u⁰(t))⁰=k(t)f^∗(u(t), t), α < t < β, u(α) = 0, u(β) =ρ.

We now claim thatu(t)≥0 for allt∈J. If the claim is false, then there exists an interval [a, b],α≤a < b≤β, such that

u(t)<0 in (a, b) and u(a) =u(b) = 0.

Consequently,

−(k(t)u⁰(t))⁰= 0, a < t < b, u(a) =u(b) = 0.

which implies thatu(t)≡0 on [a, b]. This is a contradiction and hence the claim is true. As a result, the fixed pointu∈D_σ is a nonnegative solution to (3.1).

We now put

ρ^∗= sup{ρ≥0 : (3.1) has a nonnegative solution}.

Thenρ^∗∈[σ^∗, M). Here we have used Lemma 3.4.

From the definition ofρ^∗, we can choose a sequence{ρj}^∞_j=1such thatρj < ρj+1, ρj → ρ^∗ as j 7→ +∞, and Problem (3.1) with ρJ has a nonnegative solution u_j(t) ∈ E. From Lemma 3.4 and the complete continuity of K on E, we know that {uj(t)}^∞_j=1 is uniformly bounded and equicontinuous on J. Without loss of

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generality, we may assume that uj(t)→ u^∗(t) uniformly on J as j →+∞. Note that

uj(t) = Z β

α

G(t, s)f(uj(s), s)ds+ρjh(t), t∈J . Lettingj→+∞in the above yields

u^∗(t) = Z β

α

G(t, s)f(u^∗(s), s)ds+ρ^∗h(t), t∈J.

This shows thatu^∗(t)∈E is a positive solution to (3.1) withρ^∗.

We are now in position to prove that (3.1) has at least two nonnegative solutions for allρ∈[0, ρ^∗).

For each givenρ∈[0, ρ^∗), we setξ(t)≡0 andη(t) =u^∗(t). Thenξ(t) is a lower solution to (3.1) and η(t) an upper solution. Employing the method of upper and lower solutions, we can find a solutionu(t)∈E with

0≡ξ(t)≤u(t)≤η(t) =u^∗(t) onJ .

We now assume that (H4)* holds. Using the local Lipschitz continuity of f(u, t) with respect tou∈R+ and the strong maximum principle, we deduce that

0≤u(t)< u^∗(t) for allt∈(α, β]. (3.8) If (H4) holds, i.e.,f(u, t) is nondecreasing inu∈R+ for each fixedt∈J. Then

u^∗(t)−u(t) = Z β

α

f(u^∗(s), s)−f(u(s), s)

ds+ (ρ^∗−ρ)h(t)>0 for allt∈(α, β]. i.e., (3.8) is also valid.

To obtain the existence of a second nonnegative solution to (3.1) for allρ∈[0, ρ^∗), we define a mapping ˜K:E7→E by setting

Kw˜ (t) :=

Z β α

G(t, s) ˜f(w(s), s)ds+ρh(t), and another mappingK:E×R+7→E by setting

K(w, y)(t) :=

Z β α

G(t, s)f^∗(w(s), s)ds+yh(t), where

f^∗(w, t) =

(0 ifw <0, f(w, t) ifw≥0, and

f˜(w, t) :=

(f^∗(w, t) ifw≤u^∗(t), f^∗(u^∗(t), t) ifw > u^∗(t).

Note that for each fixedw∈E, we have

kKwk˜ <kgkmax{f(u, t) : 0≤u≤ ku^∗k, t∈J}+ρ^∗=:N^∗. Pickingt₀∈(α, β) such thatu^∗(t₀)>0, we define

v^∗(t) =

(u^∗(t0) ifα≤t≤t0, u^∗(t) ift0< t≤β.

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We now put

A:=

w∈E:−M−N^∗< w(t)< v^∗(t), t∈[α, β] , B:={w∈E:kwk< M+N^∗}

where the constantM is determined by Lemma 3.4. Then bothAandB are open subsets ofE andu∈A⊂B.

Clearly, ˜K has fixed points in A. In fact, u∈Ais a fixed point of ˜K. We now consider whether ˜Khas a fixed point inB\Aor not. There are two possibilities.

Case (i). ˜K has no fixed point inB\A. In this case, we have

deg(I−K(·, ρ), A,0) = deg(I−K, A,˜ 0) = deg(I−K, B,˜ 0),

where I denotes the identity mapping from E into itself. We now claim that deg(I−K, B,˜ 0) = 1. To prove the claim, we consider the homotopic mapping

Ψ(w, τ) :=w−τKw˜ ∀(w, τ)∈E×[0,1].

For any (w, τ)∈∂B×[0,1], we have

kΨ(w, τ)k ≥ kwk − kKwk˜ > M+N^∗−N^∗=M.

i.e., Ψ(w, τ)6= 0 for any (w, τ)∈∂B×[0,1]. Consequently

deg(I−K, B,˜ 0) = deg(Ψ(·,1), B,0) = deg(Ψ(·,0), B,0) = deg(I, B,0) = 1.

On the other hand, we know that deg(I−K(·, y), B,0) is constant for all y ≥0.

From Lemma 3.4, we know that K(·, M) has no fixed point in E and hence the constant must be zero. Therefore,

deg(I−K(·, ρ), B,0) = deg(I−K(·, M), B,0) = 0.

By the excision property of the Leray-Schauder degree, we obtain deg(I−K(·, ρ), B\A,0) =−1

which implies thatK(·, ρ) has a fixed point inB\A. The fixed point is a nonnegative solution to Problem (3.1), of course.

Case (ii). ˜Khas a fixed point ¯u∈B\A. By the maximum principle, we know that 0≤u(t)¯ ≤u^∗(t) onJ.

This means that ¯u(t) is also a second solution to (3.1). Since each nonnegative solution to (3.1) is also a nonnegative radial solution to (1.2), Theorem 3.1 is thus

proved.

Finally, we consider the boundary-value problem

−∆u=fj(u), 0< α <|x|< β,

u= 0 on|x|=α, u=ρ on|x|=β, (3.9) where

f1(u) =







ξu, 0≤u≤1; 0≤ξ <1/kgk, 9(u−1)^1/9+ξ, 1≤u≤2,

η(u−2) + 9 +ξ, u≥2; η >1/m and

f2(u) :=

(sin²u, 0≤u≤8π, η(u−8π), u≥8π, η >1/m,

the constantmand the functiong(t) are determined by (3.4) and (3.2), respectively.

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Sincef1(u) satisfies (H1), (H2), (H3) and (H4),f2(u) satisfies (H1), (H2), (H3) and (H4)*, according to Theorem 3.1, there exists a positive numberρ^∗ such that Problem (3.9),j= 1, 2, has at least two nonnegative radial solutions forρ∈[0, ρ^∗), at least one forρ=ρ^∗ and none forρ > ρ^∗.

However, Theorem 1.2 cannot be applied in studying (3.9),j= 1,2.

Acknowledgement. The authors want to thank the referee for pointing out some errors in the first version of this paper.

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Wenjie Gao

Institute of Mathematics, Jilin University, Changchun 130012, China E-mail address:[email protected]

Junyu Wang

Institute of Mathematics, Jilin University, Changchun 130012, China Zhongxin Zhang

Institute of Mathematics, Jilin University, Changchun 130012, China E-mail address:[email protected]